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Let $X$ be a topological space. Suppose $A$ is clopen in $X$ and $C$ is a connected subset of $X$. If $A \cap C \neq \emptyset$, show that $C \subseteq A$.
I tried proceeding by contradiction: suppose $C\nsubseteq A$. Then this means that there is a point $x \in C$ such that $x\in A$ and $x \in X\setminus A$.
I am not sure what to do next.

wwinters57
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3 Answers3

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Since $A$ is clopen, $A$ and $X \setminus A$ form a decomposition of $X$ into two open subsets.

Each of those two sets has nonempty intersection with $C$.

So $C$ is not connected.

Lee Mosher
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If $A$ is clopen in $X$, then $X\setminus A$ is also clopen in $X$ and for any subspace $C$ of $X$, $C \cap A$ is clopen in $C$ and $C\cap (X\setminus A)= C\setminus A$ is also clopen in $C$(both by the definition of the subspace topology).

If now $C \cap A \neq \emptyset$, then we cannot have that $C\setminus A \neq \emptyset$ too, or $\{C \cap A, C\setminus A\}$ would be a non-trivial decomposition of the connected space $C$, which cannot be.

So $C\setminus A = \emptyset$, which means exactly that $C \subseteq A$..

Henno Brandsma
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Let $(X,\tau)$ be a topological space. We are told that $C$ is clopen. Thus $C$ and $X-C$ form a separation of $X$. Now if you have Munkres handy, look at lemma 23.2 of the second edition. It says that If sets $W$ and $V$ form a separation of $X$, and $Y$ is connected subspace of $X$, then $Y$ lies entirely in $W$ or $V$.

Thus in your case $A$ must lie entirely in $C$ or $X-C$. But since $A\cap C \neq \emptyset$. Thus A must lie in C

kangkan Dc
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